POJ 2182 Lost Cows

 

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10996		Accepted: 7059

Description

N (2 <= N <= 8,000) cows have unique brands in the range 1..N. In a spectacular display of poor judgment, they visited the neighborhood 'watering hole' and drank a few too many beers before dinner. When it was time to line up for their evening meal, they did not line up in the required ascending numerical order of their brands. 

Regrettably, FJ does not have a way to sort them. Furthermore, he's not very good at observing problems. Instead of writing down each cow's brand, he determined a rather silly statistic: For each cow in line, he knows the number of cows that precede that cow in line that do, in fact, have smaller brands than that cow. 

Given this data, tell FJ the exact ordering of the cows. 

Input

* Line 1: A single integer, N 

* Lines 2..N: These N-1 lines describe the number of cows that precede a given cow in line and have brands smaller than that cow. Of course, no cows precede the first cow in line, so she is not listed. Line 2 of the input describes the number of preceding cows whose brands are smaller than the cow in slot #2; line 3 describes the number of preceding cows whose brands are smaller than the cow in slot #3; and so on. 

Output

* Lines 1..N: Each of the N lines of output tells the brand of a cow in line. Line #1 of the output tells the brand of the first cow in line; line 2 tells the brand of the second cow; and so on.

Sample Input

5
1
2
1
0

Sample Output

2
4
5
3
1

Source

USACO 2003 U S Open Orange

 

另开一个数组按编号顺序记录奶牛的状态，像这样： 1 1 1 1 1

算前缀和就能知道一头牛前面的牛数量。

倒着处理数据，若最后一头牛看到前面有0头编号更小的牛，通过上面的数组就可以知道这头牛的编号是1。

记录最后一头牛的编号，然后在数组中删去这头牛，数组变成0 1 1 1 1

以此类推

 

 1 /*by SilverN*/
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 const int mxn=9000;
 9 int a[mxn];
10 int c[mxn];
11 int num[mxn];
12 int n;
13 inline int lowbit(int x){
14     return x&-x;
15 }
16 inline int sum(int x){
17     int i=x,res=0;
18     while(i){
19         res+=c[i];
20         i-=lowbit(i);
21     }
22 //    printf("sum[%d]==%d\n",x,res);
23     return res;
24 }
25 inline int find(int x){
26     int l=1,r=n;
27     int mid;
28     while(l<r){
29         mid=((l+r)>>1)+1;
30         if(sum(mid)<=x)l=mid;
31         else r=mid-1;
32     }
33     return l;
34 }
35 int main(){
36     scanf("%d",&n);
37     int i,j;
38     for(i=1;i<n;i++){
39         scanf("%d",&a[i+1]);
40     }
41     for(i=1;i<=n;i++){//初始化
42         int tmp=i;
43         while(tmp<=n){
44             c[tmp]++;
45             tmp+=lowbit(tmp);
46         }
47     }
48     for(i=n;i>=1;i--){
49         int tmp=find(a[i]+1);
50 //        printf("test %d %d\n",a[i],tmp);
51         num[i]=tmp++;
52         while(tmp<=n){
53             c[tmp]--;
54             tmp+=lowbit(tmp);
55         }
56     }
57     for(i=1;i<=n;i++)printf("%d\n",num[i]);
58     return 0;
59 }

 

转载于:https://www.cnblogs.com/SilverNebula/p/5648024.html